$f(x, y) = e^x + \cos(xy)$ $\dfrac{\partial^2 f}{\partial y \partial x} = $
Answer: Taking a mixed partial derivative is when we take two or more regular partial derivatives in a row, but each is with respect to a different variable. $\dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial x} \right]$ Let's differentiate! $\begin{aligned} \dfrac{\partial^2 f}{\partial y \partial x} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial x} \left[ e^x + \cos(xy) \right] \right] \\ \\ &= \dfrac{\partial}{\partial y} \left[ e^x - y\sin(xy) \right] \\ \\ &= 0 -\sin(xy) - xy\cos(xy) \end{aligned}$ Therefore, $\dfrac{\partial^2 f}{\partial y \partial x} = -\sin(xy) - xy\cos(xy)$.